17Calculus - Path Integrals
Path Integrals
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Let's start off with a video. This is a great video which includes a complete introduction, with examples, and explanation of applications of path integrals. It is well worth your time to watch it.
Dr Chris Tisdell - Integration over curves [45mins-37secs]
video by Dr Chris Tisdell |
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The equations from the last video are important for evaluating path integrals. Here is a summary.
requirements |
result | |
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continuous scalar function \(f(x,y,z)\) |
path integral of \(f\) over \( \mathcal{C} \) | |
curve \( \mathcal{C} \) parameterized as \(\vec{c}(t)\) on an interval \(a \leq t \leq b\) |
\(\displaystyle{ \int_{\mathcal{C}}{ f(x,y,z)~ds} = }\) \(\displaystyle{ \int_{a}^{b}{ f(\vec{c}(t)) ~ \| \vec{c}'(t) \| ~dt } }\) | |
continuous \(\vec{c}'(t)\) [derivative of \(\vec{c}(t)\) with respect to \(t\)] |
Most of the above equations should be familiar to you. However, one comment is in order about the term \(f(\vec{c}(t))\). How do you substitute a vector into a function?
If we write the vector as \( \vec{c}(t) = \langle X(t), Y(t), Z(t) \rangle = X(t) \hat{i} + Y(t) \hat{j} + Z(t)\hat{k} \) then we substitute the vector components into \(f(x,y,z)\) as \(x=X(t)\), \(y=Y(t)\) and \(z=Z(t)\). Another way to write this is
\(f(x,y,z) = f(X,Y,Z) = f(X(t), Y(t), Z(t))\). The result is a function of \(t\) only. All \(x\)'s, \(y\)'s and \(z\)'s will be gone, leaving only an integral in \(t\).
A very interesting comment he makes in the above video is that the direction of the curve does not affect the answer, i.e. integrating in the opposite direction will give the same answer. This is true because we use only the magnitude of the derivative of the parameterized curve, i.e. \(\| \vec{c}'(t) \|\), which is the same regardless of the direction. However, the path integral is dependent on the chosen path, i.e. different paths will, in general, give different results. So path integrals are sensitive to the paths.
Okay, let's try some practice problems.
Practice
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.
Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.
Solution |
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858 video
video by Dr Chris Tisdell |
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close solution
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.
Solution |
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859 video
video by Dr Chris Tisdell |
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close solution
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.
Solution |
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860 video
video by Dr Chris Tisdell |
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close solution
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Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.
Solution |
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854 video
video by Dr Chris Tisdell |
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close solution
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.
Solution |
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861 video
video by Dr Chris Tisdell |
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close solution
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Problem Statement |
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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.
Solution |
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862 video
video by PatrickJMT |
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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Problem Statement |
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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Final Answer |
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\((5\sqrt{5}+11)/6\)
Problem Statement |
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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).
Solution |
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2226 video
video by Michel vanBiezen |
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Final Answer |
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\((5\sqrt{5}+11)/6\)
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Problem Statement |
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Hint |
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There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Problem Statement |
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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).
Hint |
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There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).
Solution |
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2298 video
video by MIP4U |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
related 17calculus pages |
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external links you may find helpful |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.
